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Unformatted text preview: November 14, 2007 PHY2053 Quiz 8 (Chapter 8) Name: UFID: 1. (5 pts) A 20.0m, 40.0kg uniform ladder rests against a frictionless wall, making an angle of 60 with the horizontal. Find the minimum coefficient of static friction between ladder and ground so that a 80.0 kg firefighter can reach the top of the ladder. Since each component of the net force equals zero, we get X: nmgMg = 0 n = (m+M)g = (80+40)9.8 = 1176 N, Y: Nf = 0 f = N, where m = 80kg, M = 40kg, n/N are the normal force from the ground/the wall. Now we calculate the torque around the bottom of the ladder when the firefighter reaches the top: mgLcos +Mg(L/2)cos NLsin = 0 N = (gcos (m+M/2))/sin = 9.8cos60(80+40/2)/sin60 = 565.8 N If the coefficient is the minimum, the friction reaches the maximum static friction when...
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 Fall '06
 Buchler
 Physics, Friction

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